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strong bas titration - Practice Strong Acid/Strong Base...

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Practice Strong Acid/Strong Base Titration Calculation Problems 1. What is the pH at each of the following points in the titration of 25.00 mL of 0.50 M KOH with 0.20 M HClO 4 ? (Titration of a strong base with a strong acid) (a) before addition of HClO 4 This is just asking for the pH of a 0.50 M strong base solution so… pOH = -log(0.5) = 0.30 pH = 14 – 0.30 = 13.7 (b) after addition of 24.00 mL 0.200 M HClO 4 (before equiv point) We need to know the initial number of moles HO- as well as the number of moles of H3O+ in 24 mL of 0.200 M HClO4. Initial number of moles of HO-: 0.50 mol/L x 0.025 L = 0.0125 moles HO- Number of moles H3O+ in 24 mL of 0.200 M HClO4 0.20 mol/L x 0.024 L = 0.0048 mol H3O+ added The 0.0048 mol H3O+ will neutralize 0.0048 mol of HO- leaving (0.0125-0.0048) moles of HO- in the solution. Thus our solution has 0.0077 moles HO-. Since our new volume is (0.025 + 0.024) L = 0.049 L, our new [HO-] = 0.0077mol/0.049L= 0.157M. Thus, pOH = -log(0.157) = 0.80 So pH = 14 – 0.80 = 13.20 (c) at the equivalence point The pH at the equivalence point for a strong acid/ strong base titration is always 7.0.
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